Mathematics is the logic of certainty;

probability is the logic of uncertainty.

Certainty

x <- seq(
  from = 1, 
  to = 100, 
  length.out = 100
)

pi <- rep(
  3.1415927, 
  times = 100
)

Certainty

x <- seq(
  from = 1, 
  to = 100, 
  length.out = 100
)

pi <- rep(
  3.1415927, 
  times = 100
)

par(mar = c(5.1, 4.1, 0.01, 0.01))
plot(x, pi, type = "l")

Certainty

price <- seq(
  from = 1, 
  to = 100, 
  by = 1
)

vat = 0.25 * price

Certainty

price <- seq(
  from = 1, 
  to = 100, 
  by = 1
)

vat = 0.25 * price

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(price, vat, type = "l")

Uncertainty

Will this person buy or not?

Exercise

You work in a online retailer and are tasked with predicting if a visitor will buy the product sold or not.

In groups of 2–4 persons, discuss what sources of uncertainty exist that makes it hard to be sure about if the visitor buys or not. 5 min.

Sources of Uncertainty

• Lack of knowledge of what determines if a visitor buys.
• Lack of knowledge about the visitor.
• Measurement error of potential predictors.
• Measurement error of previous sales.
• Sampling uncertainty of past visitors.
• Randomness.

Probability of buys

The retailer has 3 visitors per day. How many of them will buy a product?

Probability of buys

1. What are all possible outcomes?

Probability of buys

2. What are we interested in knowing?

Probability of buys

2. What are we interested in knowing?

Probability of buys

3. What is the probability of the different outcomes?

Probability of buys

3. What is the probability of the different outcomes?

Probability notation

\(X\) — Random variable (r.v.)

\(\Omega_X = \{x_1, \ldots, x_n \}\) — Sample space of r.v. \(X\)

\(P(X)\) — Probability function on r.v. \(X\)

\(P(X = x_1)\) or \(P(x_1)\) — Probability of \(X\) taking value \(x_1\)

Probability distribution

We can represent random variables using probability distributions.

Binomial distribution

The retailer has 3 visitors per day. The probability of each visitor buying a product is 0.5. How many of them will buy a product?

Binomial distribution

The retailer has 3 visitors per day. The probability of each visitor buying a product is 0.5. How many of them will buy a product?

n <- 3
p <- 0.5

omega <- 0:n

prob <- dbinom(
  omega, 
  size = n, 
  prob = p
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(omega, prob, type = "h")
points(omega, prob, pch = 16)

Binomial distribution

The retailer has 10 visitors per day. The probability of each visitor buying a product is 0.5. How many of them will buy a product?

n <- 10
p <- 0.5

omega <- 0:n

prob <- dbinom(
  omega, 
  size = n, 
  prob = p
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(omega, prob, type = "h")
points(omega, prob, pch = 16)

Binomial distribution

The retailer has 10 visitors per day. The probability of each visitor buying a product is 0.3. How many of them will buy a product?

n <- 10
p <- 0.3

omega <- 0:n

prob <- dbinom(
  omega, 
  size = n, 
  prob = p
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(omega, prob, type = "h")
points(omega, prob, pch = 16)

Binomial function

Poisson distribution

Discrete duistribution for counting success in a specific time, where there is a large number of events and low probability.

Poisson distribution (PMF)

The average amount of daily sales are 4.

lambda <- 4

range <- 0:30

prob <- dpois(
  range, 
  lambda = lambda,
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(range, prob, type = "h")
points(range, prob, pch = 16)

Poisson distribution (PMF)

The average amount of daily sales are 10.

lambda <- 10

range <- 0:30

prob <- dpois(
  range, 
  lambda = lambda,
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(range, prob, type = "h")
points(range, prob, pch = 16)

Poisson distribution (CDF)

The average amount of daily sales are 4.

lambda <- 4

range <- 0:30

prob <- ppois(
  range, 
  lambda = lambda,
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(range, prob, type = "s")

Poisson distribution (CDF)

The average amount of daily sales are 4.

lambda <- 10

range <- 0:30

prob <- ppois(
  range, 
  lambda = lambda,
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(range, prob, type = "s")

Conditional probability

What is the probability that a visitor buys a product, given that it is a raining day?

Conditional probability

What is the probability that a visitor buys a product, given that it is a raining day?

Conditional probability

What is the probability that a visitor buys a product, given that it is a raining day?

Conditional probability

What is the probability that a visitor buys a product, given that it is a raining day?

Conditional probability

# Create the matrix

sun <- c(13, 39)
rain <- c(26, 104)

sales <- rbind(sun, rain)

colnames(sales) <- c("buy", "no buy")

sales

     buy no buy
sun   13     39
rain  26    104

Conditional probability

Marginal probability \(P(X = \text{Buy})\)

# P(X = Buy)
sum(sales[,"buy"]) / sum(sales)

[1] 0.2142857

# P(Y = Sun)
sum(sales["sun",]) / sum(sales)

[1] 0.2857143

Conditional probability

Conditional probability \(P(X = \text{Buy}|Y = \text{Rain})\)

# P(X = Buy|Y = Rain)
sales["rain","buy"] / sum(sales["rain", ])

[1] 0.2

# P(X = Buy|Y = Sun)
sales["sun","buy"] / sum(sales["sun", ])

[1] 0.25

sales / rowSums(sales)

      buy no buy
sun  0.25   0.75
rain 0.20   0.80

Conditional probability

Joint probability \(P(X = \text{Buy}, Y = \text{Rain})\)

sales["rain", "buy"] / sum(sales)

[1] 0.1428571

sales / sum(sales)

            buy    no buy
sun  0.07142857 0.2142857
rain 0.14285714 0.5714286

sum(sales / sum(sales))

[1] 1

Continuous random variables

Normal distribution

Normal distribution

Normal distribution

Normal distribution

Probability density function of \(\mathcal{N}(0, 1)\).

mu <- 0
sigma <- 1

x <- seq(-10, 10, by = .1)

prob <- dnorm(
  x, 
  mean = mu, 
  sd = sigma
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(x, prob, type = "l")

Normal distribution

Probability density function of \(\mathcal{N}(0, 2)\).

mu <- 0
sigma <- 2

x <- seq(-10, 10, by = .1)

prob <- dnorm(
  x, 
  mean = mu, 
  sd = sigma
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(x, prob, type = "l")

Normal distribution

Probability density function of \(\mathcal{N}(3, 2)\).

mu <- 3
sigma <- 2

x <- seq(-10, 10, by = .1)

prob <- dnorm(
  x, 
  mean = mu, 
  sd = sigma
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(x, prob, type = "l")

Normal distribution

Cumulative distribution function of \(\mathcal{N}(0, 1)\).

mu <- 0
sigma <- 1

x <- seq(-10, 10, by = .1)

prob <- pnorm(
  x, 
  mean = mu, 
  sd = sigma
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(x, prob, type = "l")

Normal distribution

Cumulative distribution function of \(\mathcal{N}(0, 2)\).

mu <- 0
sigma <- 2

x <- seq(-10, 10, by = .1)

prob <- pnorm(
  x, 
  mean = mu, 
  sd = sigma
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(x, prob, type = "l")

Normal distribution

Cumulative distribution function of \(\mathcal{N}(3, 2)\).

mu <- 3
sigma <- 2

x <- seq(-10, 10, by = .1)

prob <- pnorm(
  x, 
  mean = mu, 
  sd = sigma
)

par(mar = c(5.1, 4.1, 0.01, 0.01))

plot(x, prob, type = "l")

Normal distribution

Given \(X \sim \mathcal{N}(1, 0)\), the \(P(X \leq 1) = F_X(1)\).

mu <- 0
sigma <- 1
x <- 1

pnorm(
  x, 
  mean = mu, 
  sd = sigma
)

[1] 0.8413447

Normal distribution

Given \(X \sim \mathcal{N}(1, 0)\), calculate the probability that \(x \in [-1, 1]\).

mu <- 0
sigma <- 1
x_1 <- -1
x_2 <- 1

p_1 <- pnorm(x_1, mu, sigma)

p_2 <- pnorm(x_2, mu, sigma)

p_2 - p_1

[1] 0.6826895

Normal distribution

Simulate 1000 draws from \(X \sim \mathcal{N}(1, 0)\)

mu <- 0
sigma <- 1
n <- 1000

draws <- rnorm(n, mu, sigma)

par(mar = c(5.1, 4.1, 4.1, 0.01))

hist(draws)